tag:blogger.com,1999:blog-9079308506459446172.post5895127547548822494..comments2020-02-06T09:56:33.596-07:00Comments on SmoogeSpace: Math Proofs : Need HelpStephen Smoogenhttp://www.blogger.com/profile/17026786034163911165noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-9079308506459446172.post-22252466273829763682010-04-01T11:13:29.043-06:002010-04-01T11:13:29.043-06:00Thanks guys. I appreciate the help.Thanks guys. I appreciate the help.Stephen Smoogenhttps://www.blogger.com/profile/17026786034163911165noreply@blogger.comtag:blogger.com,1999:blog-9079308506459446172.post-8370847061529526462010-03-31T18:40:12.565-06:002010-03-31T18:40:12.565-06:00The difference between two squares is always 2n+1....The difference between two squares is always 2n+1. Since we are just looking at how far from a multiplier by 3 we are we can just mod 3 the numbers. This gives a pattern of 0,2,1,0,2,1,0,2,1...<br /><br />A number 0 (an increase by a multiplier of 3) says the next number will have the same offset as the last. A number of 1 says the offset will increase by 1. And a value of 2 says the offset will decrease by one.<br /><br />So 0, -1, +1, 0, -1, +1.... will be stuck dithering back and forth between the two places.<br /><br />The mod 5 of the sequence (0, 2, 4, 1, 3, 0, 2...) is interesting because it happens to a mod 5 number (as 0+2+4+1+3=10) just before getting the next zero, thus spending two turns at zero and explaining why two of the five numbers in the sequence are divisible by five.brejc8https://www.blogger.com/profile/18434903732348763485noreply@blogger.comtag:blogger.com,1999:blog-9079308506459446172.post-1661646202283708252010-03-31T18:14:19.392-06:002010-03-31T18:14:19.392-06:00Hi,
You have discovered modulo arithmetic :)
Bas...Hi,<br /><br />You have discovered modulo arithmetic :)<br /><br />Basically you are saying n^2 +1 != 0 mod 3<br /><br />You can just try this for small values of n:<br />n = 0 value = 0*0+1 = 1<br />n = 1 value = 1*1+1 = 2<br />n = 2 value = 2*2+1 = 5 = 2 mod 3<br /><br />The secret is any polynomial will also repeat with n mod 3. This is because you can rewrite n' as (n + 3a) where a is an integer.<br /><br />Substitute this into your original equation:<br /> (n + 3a)^2 + 1 mod 3<br />becomes:<br /> n^2 + 2 * 3a + 3a * 3a + 1 mod 3<br />becomes:<br /> n^2 + 3*(2a + 3a) + 1 mod 3<br />But mod 3 the second term is 0. So:<br /> n^2 + 1 mod 3<br /><br />Hence the repeating 1 2 2 results for n mod 3<br /><br />Regards,<br /><br />DDD<br /><br />P.S The maths was way easier than logging into this comments system :)DDDhttps://www.blogger.com/profile/08396087485745915062noreply@blogger.comtag:blogger.com,1999:blog-9079308506459446172.post-16432642190751182392010-03-31T18:06:19.036-06:002010-03-31T18:06:19.036-06:00Its all about the modulus of 3 and the quadratic e...Its all about the modulus of 3 and the quadratic equation.<br /><br />Take n=a*3+b where a is any integer<br /> take b as 0, 1 or 2<br /><br />Now n^2=9*a^2+6*a*b+b^2=3*C+b^2<br /><br />b^2 is either 0, 1, 4<br />4=3+1<br /><br />when b=0 n^2=3c<br />when b=1 n^2=3c+1 <br />when b=2 n^2=3c+1<br /> <br />Now m=n^2+1=3*c+b^2+1<br />when b=0 n^2+1=3c+1<br />when b=1 n^2+1=3c+2<br />when b=2 n^2+1=3c+2Jef Spaletahttps://www.blogger.com/profile/11439754449677675460noreply@blogger.comtag:blogger.com,1999:blog-9079308506459446172.post-51552460746828396292010-03-31T17:56:51.416-06:002010-03-31T17:56:51.416-06:00The fact that (n^2+1) is never divisible by 3 can ...The fact that (n^2+1) is never divisible by 3 can be proven like this:<br /><br />Two small reminders:<br />* "X is divisible by 3" <=> X % 3 == 0 (using the C notation)<br />* (A * B) % 3 == (((A % 3) * (B % 3)) % 3)<br /><br />Therefore, (n^2)%3 == ((n%3)^2)%3, and we have three cases to consider:<br />* n%3 == 0, then (n^2)%3 == 0<br />* n%3 == 1, then (n^2)%3 == 1<br />* n%3 == 2, then (n^2)%3 == 4%3 == 1<br /><br />We see that (n^2)%3 != 2, therefore (n^2+1)%3 != 0, hence (n^2+1) is never divisible by 3.<br /><br />As for "diff between m", start with differences between n^2:<br />* (n+1)^2 == n^2 + (2n+1)<br /><br />Your "diff between m" is thus (2n+1)/3, except for rounding. Similarly to the above, you can determine the structure of (2n+1)%3. Looking at this will reveal the reason why two out of three values are repeated, details left as an exercise :)Miloslavhttps://www.blogger.com/profile/12507179914843206625noreply@blogger.comtag:blogger.com,1999:blog-9079308506459446172.post-49881672119822837922010-03-31T17:36:57.348-06:002010-03-31T17:36:57.348-06:00n^2 + 1 = 3m would mean that 2 is a quadratic resi...n^2 + 1 = 3m would mean that 2 is a quadratic residue mod 3. But there are only (p - 1)/2 quadratic residues for an odd prime modulus. For 3, the sole quadratic residue is 1...<br /><br />To see why, look at the field Z/pZ. The squares 1^2, 2^2,..., (p-1)/2 ^2 enumerate all the quadratic residues mod p. They are all different, since x^2 = y^2 implies (x + y)(x - y) = 0, ie x must be y or -y.Unknownhttps://www.blogger.com/profile/08624433288530789872noreply@blogger.com